You have found the following ages (in years) of all 5 lions at your local zoo: $ 3,\enspace 5,\enspace 2,\enspace 9,\enspace 10$ What is the average age of the lions at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Explanation: Because we have data for all 5 lions at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $5$ ages and divide by $5$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{5}} x_i}{{5}} $ $ {\mu} = \dfrac{3 + 5 + 2 + 9 + 10}{{5}} = {5.8\text{ years old}} $ Find the squared deviations from the mean for each lion. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $3$ years $-2.8$ years $7.84$ years $^2$ $5$ years $-0.8$ years $0.64$ years $^2$ $2$ years $-3.8$ years $14.44$ years $^2$ $9$ years $3.2$ years $10.24$ years $^2$ $10$ years $4.2$ years $17.64$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{7.84} + {0.64} + {14.44} + {10.24} + {17.64}} {{5}} $ $ {\sigma^2} = \dfrac{{50.8}}{{5}} = {10.16\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{10.16\text{ years}^2}} = {3.2\text{ years}} $ The average lion at the zoo is 5.8 years old. There is a standard deviation of 3.2 years.